LeetCode Remove Nth Node From End of List

原题链接在这里：https://leetcode.com/problems/remove-nth-node-from-end-of-list/

Method 1: 算出总长度，再减去n，即为要从头多动的点。但要新要求，only one pass。

Method 2: 两个指针，一个runner，一个walker，runner先走n步，随后runner和walker一起走，直到runner指为空。

Note: 1. 都是找到要去掉点的前一个点记为mark，再通过mark.next = mark.next.next去掉对应应去掉的点。

2. 注意去掉list头点的情况，e.g. 1->2, n = 2.

AC Java:

/**
* Definition for singly-linked list.
* public class ListNode {
*     int val;
*     ListNode next;
*     ListNode(int x) { val = x; }
* }
*/
public class Solution {
public ListNode removeNthFromEnd(ListNode head, int n) {
/*Method 1
if(head == null)
return head;
int len = 0;
ListNode temp = head;
while(temp != null){
temp = temp.next;
len++;
}
if(len < n){
return head;
}
int moveNum = len - n;
ListNode dunmy = new ListNode(0);
dunmy.next = head;
temp = dunmy;
while(moveNum > 0){
temp = temp.next;
moveNum--;
}
temp.next = temp.next.next;
return dunmy.next;
*/

//Method 2
if(head == null || n == 0)
return head;
ListNode dunmy = new ListNode(0);
dunmy.next = head;
ListNode runner = dunmy.next;
ListNode walker = dunmy;
while(n>0 && runner != null){
runner = runner.next;
n--;
}
while(runner != null){
runner = runner.next;
walker = walker.next;
}
walker.next = walker.next.next;
return dunmy.next;
}
}